Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define LL long long
#define pii pair<int,int>
using namespace std;
const int inf = 2147483647;
const int N = 200001;

int _max(int x, int y) {return x > y ? x : y;}
int _min(int x, int y) {return x < y ? x : y;}
inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
    return x * f;
}
void put(int x) {
	if(x < 0) putchar('-'), x = -x;
	if(x >= 10) put(x / 10);
	putchar(x % 10 + '0');
}

int p[N], f[N], T[N];

int check(int a, int b, int c) {
	return (a < b && b < c) || (b < c && c < a) || (c < a && a < b);
}

int main() {
	int n = read();
	for(int i = 1; i <= n; i++) p[read()] = i;
	for(int i = n - 1; i; i--) {
		if(!f[i + 1]) {
			if(p[i] > p[i + 1]) T[i] = 1, f[i] = i + 1;
		} else {
			if(check(p[i], p[i + 1], p[f[i + 1]])) T[i] = T[i + 1], f[i] = f[i + 1];
			else T[i] = T[i + 1] + 1, f[i] = i + 1;
		}
	} put(T[1]), puts("");
	return 0;
}